Added Urysohn Metrisation theorem and compactness theorems.

src/topology/main/regular.tex

@@ -37,3 +37,25 @@
 
     as $V$ is closed. Therefore $F(U) \subset V$, and $F$ is continuous.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:second-countable-regular}
+    Let $X$ be a second countable regular space, then $X$ is normal. 
+\end{proposition}
+\begin{proof}
+    Let $\cb \subset 2^X$ be a base for $X$ and $A, B \subset X$ be disjoint closed sets. For each $x \in A$ and $y \in B$, there exists $U_x, V_y \in \cb$ such that $x \in \ol{U_x} \subset B^c$ and $y \in \ol{V_y} \subset B^c$. Since $\cb$ is countable, let $\seq{U_n}$ and $\seq{V_n}$ be enumerations of $\bracs{U_x|x \in A}$ and $\bracs{V_y|y \in B}$, respectively. 
+
+    For each $n \in \natp$, let
+    \[
+    U_n' = U_n \setminus \bigcup_{j = 1}^n \ol{V_j} \quad V_n' = V_n \setminus \bigcup_{j = 1}^n \ol{V_j}
+    \]
+
+    then $U_n'$ and $V_n'$ are both open. Let
+    \[
+    U = \bigcup_{n \in \natp}U_n' \quad V = \bigcup_{n \in \natp}V_n'
+    \]
+
+    then $U \in \cn_X(A)$ and $V \in \cn_X(B)$. For each $m, n \in \natp$ with $m \le n$, $V_n \cap \bigcup_{j = 1}^n U_j = \emptyset$, so $U_m \cap V_n = \emptyset$. Likewise, if $m \ge n$, then $U_m \cap V_n = \emptyset$ as well. Therefore $U \cap V = \emptyset$. 
+\end{proof}
+
+